Problem: Divide the following complex numbers. $ \dfrac{2-16i}{-3-i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3+i}$ $ \dfrac{2-16i}{-3-i} = \dfrac{2-16i}{-3-i} \cdot \dfrac{{-3+i}}{{-3+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(2-16i) \cdot (-3+i)} {(-3-i) \cdot (-3+i)} = \dfrac{(2-16i) \cdot (-3+i)} {(-3)^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(2-16i) \cdot (-3+i)} {(-3)^2 - (-1i)^2} = $ $ \dfrac{(2-16i) \cdot (-3+i)} {9 + 1} = $ $ \dfrac{(2-16i) \cdot (-3+i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({2-16i}) \cdot ({-3+i})} {10} = $ $ \dfrac{{2} \cdot {(-3)} + {-16} \cdot {(-3) i} + {2} \cdot {1 i} + {-16} \cdot {1 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{-6 + 48i + 2i - 16 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{-6 + 48i + 2i + 16} {10} = \dfrac{10 + 50i} {10} = 1+5i $